Đáp án:
$\begin{array}{l}
c)\\
C = \left( {4\sqrt 2 + 1} \right)\left( {3 + 2\sqrt {2 + \sqrt 2 } } \right)\left( {3 - 2\sqrt {2 + \sqrt 2 } } \right)\\
= \left( {4\sqrt 2 + 1} \right)\left( {{3^2} - 4.\left( {2 + \sqrt 2 } \right)} \right)\\
= \left( {4\sqrt 2 + 1} \right)\left( {1 - 4\sqrt 2 } \right)\\
= 1 - {\left( {4\sqrt 2 } \right)^2}\\
= 1 - 32\\
= - 31\\
d)D = \sqrt {10 + 2\sqrt {3 - 2\sqrt {29 - 12\sqrt 5 } } } \\
= \sqrt {10 + 2\sqrt {3 - 2\sqrt {20 - 2.2\sqrt 5 .3 + 9} } } \\
= \sqrt {10 + 2\sqrt {3 - 2\sqrt {{{\left( {2\sqrt 5 - 3} \right)}^2}} } } \\
= \sqrt {10 + 2\sqrt {3 - 2\left( {2\sqrt 5 - 3} \right)} } \\
= \sqrt {10 + 2\sqrt {9 - 4\sqrt 5 } } \\
= \sqrt {10 + 2\sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} } \\
= \sqrt {10 + 2\left( {\sqrt 5 - 2} \right)} \\
= \sqrt {6 + 2\sqrt 5 } \\
= \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} \\
= \sqrt 5 + 1
\end{array}$
